Answer to Question #179624 in Statistics and Probability for Ajay

Question

A factory produces steel pipes in three plant with daily production volumes of 500, 

1000 and 2000 units respectively from each of the plants. From the past experience it 

is known that the fraction of defective outputs produced by three plants are 

respectively 0.005, 0.008 and 0.010. If a pipe is selected at random from a day’s total 

production and founded to be defective, from which plant is that likely to 

have came?

Answer 1

Solution:

Given that daily production volumes of 500, 1000 and 2000 units respectively.

The fraction of defective outputs produced by three plants are 

respectively 0.005, 0.008 and 0.010.

So, number of defective outputs by first plant = 500 x 0.005 = 2.5

Number of defective outputs by second plant = 1000 x 0.008 = 8

Number of defective outputs by third plant = 2000 x 0.010 = 20

Clearly, third plant produced greater number of defective outputs, so a selected pipe is likely to come from this plant.

Hence, third plant is the required answer.